设数列{An}是首项为a1,a1>0,公差为2地等差数列,其前N项和为Sn,且根号S1,根号S2,根号S3成等差数列.求
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设数列{An}是首项为a1,a1>0,公差为2地等差数列,其前N项和为Sn,且根号S1,根号S2,根号S3成等差数列.求数列{an}的通项公式,(2)记Bn=an/2^n的前n项和为Tn,求Tn
1)an=a1+2(n-1),Sn=na1+n(n-1)
a1=1,an=2n-1(n=1,2,3...)
2)Bn=(2n-1)/2^n=1/2+3/4+5/8+7/16+...+(2n-1)/2^n ①
Bn*1/2, Bn/2= 0 +1/4+3/8+5/16+...+(2n-1)/2^(n+1) ②
①-②得,Bn/2=1/2+(1/2+1/4+
a1=1,an=2n-1(n=1,2,3...)
2)Bn=(2n-1)/2^n=1/2+3/4+5/8+7/16+...+(2n-1)/2^n ①
Bn*1/2, Bn/2= 0 +1/4+3/8+5/16+...+(2n-1)/2^(n+1) ②
①-②得,Bn/2=1/2+(1/2+1/4+
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